Proper and dense subgroup of $\mathrm{SU}(2)$

Question

Question : does there exist a proper and dense subgroup of $\mathrm{SU}(2)$ ?

An idea of such a group would be the following. Consider the following basis of $\mathfrak{su}(2):=\mathrm{Lie}(\mathrm{SU}(2))$ : $$ i\sigma_1 = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \qquad i\sigma_2 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \qquad i\sigma_3 = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} $$ Let $a_1,a_2,a_3\in \mathbb Q^c\cap ]0,1[$ be three irrational numbers between 0 and 1. Let $$ g_1 := \exp(2\pi i a_1 \sigma_1) \qquad g_2 := \exp(2\pi i a_1 \sigma_2) \qquad g_3 := \exp(2\pi i a_3 \sigma_3) $$ be three elements of $\mathrm{SU}(2)$. Consider $H<\mathrm{SU}(2)$ be the subgroup of $\mathrm{SU}(2)$ generated by $g_1,g_2,g_3$. We know that $H$ is proper in $\mathrm{SU}(2)$ because it is countable (and it is countable because it is finitely generated). But I don't know if $H$ is dense in $\mathrm{SU}(2)$.

Answer 1

Consider the closure of $H$. A closed subgroup of a Lie group is a Lie subgroup. The subgroup $\overline H$ contains each $\exp(tA)$ for $A=i\sigma_j$ in your notation, so its Lie algebra contains all of the $i\sigma_j$, and so is all of $\frak su(2)$. Thus $\overline H$ must be $SU(2)$.

Answer 2

Late to the party, but a possibly interesting little number-theoretic point: for any quaternion algebra $B$ over $\mathbb Q$, with $B\otimes_{\mathbb Q}\mathbb R$ isomorphic to the Hamiltonian quaternions, the norm-one elements of $B$ have a natural map to $SU(2)$, and have dense image (for various reasons). (Further details are not expensive, if they'd be of interest.)